Further Improvised Question: Combination of selection of pens Following from my first improvised question here and the two excellent answers given, here's another twist to the question. What happens if the total number of pens to be selected is $15$ instead of $10$? 
The rephrased question would as follows:

Supposing there are four brands of pens, W, X, Y, Z. You want to choose $\color{red}{15}$ pens made up of any combination of the brands, but limited to a maximum of $5$ pens from each brand. How many possible combinations are there? Assume it would be acceptable not to choose a pen from any one or more brands, and that pens from any one brand are indistinguishable from each other. 

This is different from the previous question in that in the previous question, once the $5$ pens have been selected from one brand, none of the other brands will hit the limit imposed. But this is not the case here, so it would be slightly more complex. Who would have thought that choosing pens is such a complex activity!
 A: In the previous link, for generating inappropriate selections to subtract, it was possible to pre-select 6 pens in any one of the types. Now we can also do so in upto 2 types, so subtract all possible inappropriate selections using inclusion-exclusion.
Unrestricted selections using stars and bars  will now be ${15+4-1\choose 4-1} = {18\choose 3}$, and restricted ones,
$${18\choose3} - {4\choose 1}{12\choose 3} +{4\choose 2}{6\choose 3} = 56$$    
A: Two approaches:
One is imagine there are $5$ pens of each brand making $20$ in total, and you want to find the ways of not selecting $20-15=5$ of them.  Using the methods in answers to your two earlier questions, this is $\binom{8}{3}=56$.
Another is to look at the generating function $1+x+x^2+x^3+x^4+x^5=\dfrac{1-x^6}{1-x}$ for the number of ways of choosing $n$ pens from five of the same brand, taking the fourth power to take account of the number of brands, and looking at the coefficient of $x^{15}$ since you are interested in $15$ pens in total.  It gives the same answer, as the full expansion is $${{x}^{20}}+4 {{x}^{19}}+10 {{x}^{18}}+20 {{x}^{17}}+35 {{x}^{16}}+56 {{x}^{15}}+80 {{x}^{14}}+104 {{x}^{13}}+125 {{x}^{12}}+140 {{x}^{11}}+146 {{x}^{10}}+140{{x}^{9}}+125{{x}^{8}}+104{{x}^{7}}+80 {{x}^{6}}+56{{x}^{5}}+35 {{x}^{4}}+20 {{x}^{3}}+10 {{x}^{2}}+4 x+1$$ 
